Corrected exercises: geometric optics and optical instruments (203-SN3-RE)

$\sin\theta_{2}=\frac{n_{1}\sin\theta_{1}}{n_{2}}=\frac{\sin 50.0°}{1.33}\approx 0.5760$, so $\theta_{2}\approx 35.2°$: the ray bends toward the normal on entering the optically denser medium. Speed: $v=\frac{c}{n}=\frac{3.00\times 10^{8}}{1.33}\approx 2.26\times 10^{8}$ m/s. The FREQUENCY does not change: it is set by the source and nothing can alter it along the way (the atoms at the interface vibrate at the rhythm of the incident wave). Since $\lambda=\frac{v}{f}$, it is the WAVELENGTH that gets divided by $n$: the light bunches up as it slows down.

For nearly vertical rays the angles are small and $\sin\theta\approx\tan\theta\approx\theta$. A ray leaving the coin at angle $\theta_{1}$ in the water emerges at $\theta_{2}$ with $n\theta_{1}\approx\theta_{2}$. It reaches the surface at a horizontal distance $x\approx d\theta_{1}$ from the point directly above the coin, and its backward extension meets the vertical at the depth $d'\approx\frac{x}{\theta_{2}}=\frac{d\theta_{1}}{n\theta_{1}}=\frac{d}{n}$. Numerically: $d'=\frac{2.40}{1.33}\approx 1.80$ m. The pool looks a quarter shallower than it really is, which explains many unpleasant surprises when diving in.

$\sin\theta_{c}=\frac{n_{air}}{n_{water}}=\frac{1}{1.33}\approx 0.7519$, so $\theta_{c}\approx 48.8°$. This is Snell's window: everything outside, from the horizon to the zenith, is compressed into a cone of half-angle 48.8° above the fish. Outside this cone no light from the outside world can reach it: the surface then behaves like a perfect MIRROR, and the fish sees there the reflection of the bottom and of everything around it underwater.

At entry, $\sin 50.0°=1.50\sin r$ gives $r\approx 30.7°$. At exit, the second face being parallel to the first, the internal angle of incidence is again $r$ and $1.50\sin r=\sin\theta_{out}$, which is exactly the starting relation read backward: $\theta_{out}=50.0°$. The emerging ray is therefore strictly parallel to the incident ray; the slab only SHIFTS it. Displacement: $d=e\frac{\sin(i-r)}{\cos r}=(0.0200)\frac{\sin(19.3°)}{\cos(30.7°)}\approx 7.68\times 10^{-3}$ m, that is 7.7 mm. This is why a windowpane does not distort an image but shifts it slightly, more so when it is thick and viewed obliquely.

$\sin\theta_{c}=\frac{1.45}{1.50}\approx 0.9667$, so $\theta_{c}\approx 75.2°$: the ray must strike the wall beyond this angle, that is, travel within $14.8°$ of the axis. Numerical aperture: $NA=\sqrt{1.50^{2}-1.45^{2}}=\sqrt{0.1475}\approx 0.384$, and the acceptance half-angle in air is $\arcsin(0.384)\approx 22.6°$: only light entering within this cone stays trapped. As for Snell's law, it is not a decree: it follows from Fermat's principle, according to which light follows the path of stationary TRAVEL TIME between two points. Going straight would not be fastest, since the second medium is slower: the optimal path shortens the slow portion at the price of a detour in the fast portion, exactly like a lifeguard who runs farther along the sand in order to swim less. Writing down this condition and differentiating gives back $n_{1}\sin\theta_{1}=n_{2}\sin\theta_{2}$, to the letter.

$\dfrac{1}{d_{i}}=\dfrac{1}{f}-\dfrac{1}{d_{o}}=\dfrac{1}{20.0}-\dfrac{1}{30.0}=\dfrac{3-2}{60}=\dfrac{1}{60}$, so $d_{i}=+60.0$ cm. The magnification is $m=-\dfrac{60.0}{30.0}=-2.00$. The image is therefore REAL since $d_{i}>0$, INVERTED since $m<0$, and ENLARGED twice since $\left|m\right|=2$. It forms 60 cm on the other side of the lens and could be caught on a screen: this is the projector configuration.

$\dfrac{1}{d_{i}}=\dfrac{1}{20.0}-\dfrac{1}{15.0}=\dfrac{3-4}{60}=-\dfrac{1}{60}$, so $d_{i}=-60.0$ cm and $m=-\dfrac{-60.0}{15.0}=+4.00$. The image is VIRTUAL, $d_{i}<0$, UPRIGHT, $m>0$, and enlarged four times. It lies on the same side as the object and cannot be projected: you can only look at it through the lens. This is exactly how a MAGNIFYING GLASS works.

$\dfrac{1}{d_{i}}=\dfrac{1}{20.0}-\dfrac{1}{60.0}=\dfrac{3-1}{60}=\dfrac{2}{60}$, so $d_{i}=+30.0$ cm and $m=-\dfrac{30.0}{60.0}=-0.500$. The image is real, inverted and REDUCED by half. This is the camera configuration, in which a wide scene is brought down onto a small sensor.

$\dfrac{1}{d_{i}}=\dfrac{1}{-20.0}-\dfrac{1}{30.0}=\dfrac{-3-2}{60}=-\dfrac{5}{60}$, so $d_{i}=-12.0$ cm and $m=-\dfrac{-12.0}{30.0}=+0.400$. The image is VIRTUAL, UPRIGHT and REDUCED. This is the case for ANY diverging lens, wherever the object is: it can never form a real image of a real object.

For a converging lens, everything depends on the position of the object relative to the focal point. If $d_{o}>f$, the object is beyond the focal point and the image is REAL and inverted; it is enlarged if $f<d_{o}<2f$, the same size if $d_{o}=2f$, reduced if $d_{o}>2f$. If $d_{o}<f$, the object is inside the focal point and the image becomes VIRTUAL, upright and enlarged. The borderline case $d_{o}=f$ gives no image: the rays emerge parallel and the image is pushed out to infinity, which is what $\dfrac{1}{d_{i}}=0$ expresses.

Ray 1, parallel to the axis, emerges through F'; ray 2, which passes through F, emerges parallel to the axis; ray 3 goes through the centre O undeviated. The three meet 40.0 cm behind the lens, exactly on 2F'. The calculation confirms it: $\dfrac{1}{d_{i}}=\dfrac{1}{20.0}-\dfrac{1}{40.0}=\dfrac{1}{40.0}$, so $d_{i}=+40.0$ cm and $m=-1.00$: the image is real, inverted and the same size as the object. It is the boundary between the enlarged image of a), object between 2F and F, and the reduced image of c), object beyond 2F. The ray diagram is also the best check on a calculation: two rays are enough to place the image, and the third must pass through the same point, otherwise an error has crept in somewhere.

$f=\dfrac{R}{2}=\dfrac{40.0}{2}=+20.0$ cm. The sign is POSITIVE because the mirror is concave, hence converging: it brings a parallel beam arriving from the front to a real focal point. The convention is consistent with that of lenses, where $f>0$ means converging.

$\dfrac{1}{d_{i}}=\dfrac{1}{20.0}-\dfrac{1}{30.0}=\dfrac{1}{60}$, so $d_{i}=+60.0$ cm and $m=-2.00$. The image is REAL, formed 60 cm IN FRONT of the mirror, on the same side as the object since the light comes back, INVERTED and enlarged twice. The numbers are identical to those of part a) of exercise 2, which shows that the formula does not distinguish a lens from a mirror; only the geometric reading of “on the same side” changes.

For a convex mirror, $R$ is counted negatively, so $f=-20.0$ cm. Then $\dfrac{1}{d_{i}}=\dfrac{1}{-20.0}-\dfrac{1}{30.0}=-\dfrac{5}{60}$, so $d_{i}=-12.0$ cm and $m=+0.400$. The image is VIRTUAL, located behind the mirror, UPRIGHT and reduced. This is the invariable behaviour of the convex mirror, exactly like the diverging lens.

Because a convex mirror always gives an upright, REDUCED image, which widens the field of view considerably: you see a much wider stretch of road than with a flat mirror of the same size, which shrinks the blind spot. The price to pay is precisely the reduction: the vehicle behind looks smaller, so the brain, which estimates distances from apparent size, judges it FARTHER than it really is. Hence the engraved warning, mandatory in North America.

The object must be placed BETWEEN the mirror and its focal point, so closer than 20.0 cm here. As with the converging lens inside its focal length, you then get a virtual, upright, enlarged image. Common uses: the makeup or shaving mirror, and the dentist's examination mirror, both concave and used very close to the face. Check: with $d_{o}=10.0$ cm, $\dfrac{1}{d_{i}}=\dfrac{1}{20}-\dfrac{1}{10}=-\dfrac{1}{20}$, so $d_{i}=-20.0$ cm and $m=+2.00$, a virtual upright image enlarged twice.

Ray 1, parallel to the axis, reflects through F; ray 2, which passes through F, reflects parallel to the axis; ray 3, which passes through the centre of curvature C, strikes the mirror at a right angle and returns along itself. The three meet BETWEEN C and F, in front of the mirror. The calculation confirms it: $\dfrac{1}{d_{i}}=\dfrac{1}{20.0}-\dfrac{1}{60.0}=\dfrac{2}{60}$, so $d_{i}=+30.0$ cm and $m=-0.500$: real image, inverted, reduced by half, 30 cm in front of the mirror. This is b) read backward: object and image have swapped places, because the equation is symmetric in $d_{o}$ and $d_{i}$ and light can always retrace the same path in the opposite direction.

The FAR POINT is the most DISTANT point seen sharply, with the eye relaxed; it is at INFINITY for a normal eye. The NEAR POINT is the CLOSEST point seen sharply, with the eye accommodating to the maximum; by convention it is at 25 cm. Between the two lies the range of clear vision, which accommodation, that is, the reshaping of the crystalline lens, allows the eye to cover.

The nearsighted eye cannot see beyond 50.0 cm. The lens must therefore take an object at INFINITY and give a virtual image located precisely at the far point, where the eye can see it. So $d_{o}=\infty$ and $d_{i}=-0.500$ m, the sign being negative because the image is virtual, on the same side as the object. Then $\dfrac{1}{f}=\dfrac{1}{d_{o}}+\dfrac{1}{d_{i}}=0+\dfrac{1}{-0.500}=-2.00$. The power is $-2.00$ dioptres: a DIVERGING lens.

The farsighted eye cannot focus on anything closer than 100 cm. To read at 25.0 cm, the lens must take the real object at $d_{o}=+0.250$ m and form a virtual image at the near point, that is $d_{i}=-1.00$ m. Then $\dfrac{1}{f}=\dfrac{1}{0.250}+\dfrac{1}{-1.00}=4.00-1.00=3.00$. The power is $+3.00$ dioptres: a CONVERGING lens.

Because the two defects are opposite. The NEARSIGHTED eye converges TOO MUCH: the image of a distant object forms in front of the retina. Some convergence must therefore be taken away, hence a diverging lens of negative power. The FARSIGHTED eye does NOT converge ENOUGH: the image would form behind the retina. Some must be added, hence a converging lens of positive power. Since the powers of two lenses in contact add, the correction literally consists of adding algebraically what is missing or what is in excess.

Because presbyopia is not a defect in the shape of the eye but a loss of ACCOMMODATION: the crystalline lens hardens with age and can no longer bulge as much, which gradually pushes the near point away. The phenomenon is physiological and so affects everyone, including nearsighted people, in whom it adds to the existing defect. The consequence, a near point that is too far, is the SAME as in farsightedness, and the correction is therefore the same: a converging lens for near vision. This is why a presbyopic nearsighted person needs two opposite corrections, hence bifocal or progressive lenses.

Because the apparent size of an object depends not on its real size but on the ANGLE it subtends at the eye. An object twice as large but twice as far looks identical. A magnifier does not change the object; it lets you observe it from much closer, a few centimetres away instead of 25, by forming a virtual image that the eye can accommodate. It is the increase in angle that produces the sensation of enlargement, hence the definition $M=\dfrac{\theta'}{\theta}$.

$M=\dfrac{25}{f}=\dfrac{25}{5.00}=5.00$. The magnifier therefore magnifies five times. The formula assumes the image is thrown out to infinity, the most restful configuration for the eye, which then does not accommodate at all.

The object must be placed BETWEEN the magnifier and its focal point, so closer than 5.00 cm, and in practice almost exactly at the focal point. The image obtained is then VIRTUAL, UPRIGHT and enlarged, in agreement with case b) of exercise 2. It has to be a virtual image: a real image would be inverted and would have to be projected, which is not how a magnifier is used.

$M=\dfrac{25}{f}=10$ gives $f=2.5$ cm. The drawback is immediate: the object must be brought to 2.5 cm from the lens, and the eye must sit right against it. The field of view becomes tiny, lighting is difficult because the head casts a shadow, and the depth of field is almost nil. The stronger the magnification, the more uncomfortable the magnifier becomes.

It would take $f=\dfrac{25}{100}=0.25$ cm, that is 2.5 mm. A lens of this focal length would be a glass bead a few millimetres across, with enormous spherical and chromatic aberrations, a field of the order of a millimetre, and a working distance incompatible with any handling. Beyond a magnification of about 20, one therefore stops adding curvature to a single lens and moves to a TWO-stage system: this is exactly the reason the microscope exists, the subject of a later exercise.

$\sin 45.0°=1.50\sin r_{1}$ gives $\sin r_{1}=\frac{0.7071}{1.50}\approx 0.4714$, so $r_{1}\approx 28.1°$. Then $r_{2}=60.0-28.1=31.9°$, and $\sin i_{2}=1.50\sin 31.9°\approx 0.7919$, that is $i_{2}\approx 52.4°$. Deviation: $\delta=45.0+52.4-60.0\approx 37.4°$.

At the minimum, $r_{1}=r_{2}=30.0°$, so $\sin i=1.50\sin 30.0°=0.750$ and $i\approx 48.6°$. Minimum deviation: $\delta_{min}=2i-A=97.18-60.0\approx 37.2°$. Compare: when the incidence changes from 48.6° to 45.0°, a change of more than three and a half degrees, the deviation has moved by only two tenths of a degree. This is the signature of a minimum: near an extremum, the function is FLAT. This insensitivity is precisely what makes the minimum deviation easy to find experimentally, by eye, by rotating the prism until the image stops moving back and starts moving the other way.

At the minimum, $i=\frac{A+\delta_{min}}{2}=\frac{60.0+37.18}{2}\approx 48.59°$ and $r=\frac{A}{2}=30.0°$. Snell's law at entry reads $\sin i=n\sin r$, so $n=\frac{\sin\frac{A+\delta_{min}}{2}}{\sin\frac{A}{2}}=\frac{\sin 48.59°}{\sin 30.0°}=\frac{0.750}{0.500}=1.50$: the starting index, recovered. It is the reference method because it needs no distance measurement, no image position and no delicate adjustment: only TWO ANGLES, $A$ and $\delta_{min}$, both measurable on a goniometer to a hundredth of a degree and located unambiguously thanks to the flatness of the minimum. An index can thus be measured to five significant figures with a purely mechanical instrument.

For the ray to exit, its internal incidence on the second face must stay BELOW the critical angle: $r_{2}\leq\theta_{c}$ with $\sin\theta_{c}=\frac{1}{1.50}$, that is $\theta_{c}\approx 41.8°$. Since $r_{2}=A-r_{1}$, this requires $r_{1}\geq 60.0-41.8=18.2°$, so $\sin i_{1}\geq 1.50\sin 18.2°\approx 0.4683$ and $i_{1}\geq 27.9°$. Below 27.9°, the light undergoes total internal reflection inside the prism and leaves through the third face, or not at all: this is the very phenomenon exploited by the total-reflection prisms of binoculars, which replace mirrors with no loss at all.

Red: $i=\arcsin(1.495\times 0.500)$, so $\delta_{min,r}=2\arcsin(0.7475)-60.0\approx 36.75°$. Violet: $\delta_{min,v}=2\arcsin(0.7575)-60.0\approx 38.49°$. Difference: about $1.74°$, nearly two degrees between the two ends of the visible spectrum, plenty to spread a rainbow across a wall. The difference with the parallel-sided slab is purely geometric: in the slab, the two faces are parallel, so the second refraction exactly CANCELS the first for each colour, and all the rays emerge parallel to one another, separated only by a tiny lateral shift that makes them overlap right away. In the prism, the two faces make an angle: the two deviations ADD instead of compensating, and each colour leaves in its own direction, which drifts further and further from the others as you move away.

$\frac{1.00}{20.0}+\frac{1.50}{d_{i}}=\frac{1.50-1.00}{4.00}=0.125$. So $\frac{1.50}{d_{i}}=0.125-0.0500=0.0750$ and $d_{i}=\frac{1.50}{0.0750}=20.0$ cm: the image forms 20.0 cm inside the glass, and it is REAL ($d_{i}>0$). Note that the usual thin lens equation is just the special case $n_{1}=n_{2}=1$ of this one, applied twice.

The entry face is convex toward the incident light, so $R_{1}=+20.0$ cm; the exit face bulges the other way, so its centre is on the side of the INCOMING light and $R_{2}=-30.0$ cm. $\frac{1}{f}=(1.52-1)\left(\frac{1}{20.0}+\frac{1}{30.0}\right)=0.52\times 0.08333\approx 0.04333$ cm$^{-1}$, so $f\approx 23.1$ cm and $P=\frac{1}{0.231}\approx 4.33$ D.

$\frac{n}{n_{water}}-1=\frac{1.52}{1.33}-1\approx 0.1429$, almost four times less than 0.52. The focal length is multiplied accordingly: $f\approx 84.0$ cm, against 23.1 cm in air. The lesson applies to the eye too: its cornea provides two thirds of the eye's converging power precisely because it separates AIR from watery media; underwater, this jump in index nearly disappears, the cornea stops converging and the image forms far behind the retina, hence the total blur. A diving mask restores an air-cornea interface by trapping a layer of air in front of the eye: it is not the mask that corrects the vision, it is the air it contains.

The two faces facing the same way have radii of the SAME sign: $R_{1}=+10.0$ cm and $R_{2}=+15.0$ cm. $\frac{1}{f}=0.52\left(\frac{1}{10.0}-\frac{1}{15.0}\right)=0.52\times 0.03333\approx 0.01733$, so $f\approx 57.7$ cm, POSITIVE: the lens is indeed converging, despite its hollow face. It is not the shape of each face taken separately that decides, but the DIFFERENCE between their curvatures, in other words the fact that the lens is thicker at the centre than at the edges. Incidentally, all modern spectacle lenses are menisci, for reasons of field of view and appearance.

Two thin lenses in contact: the image of the first serves as the object of the second, and since the gap between them is zero, $d_{o2}=-d_{i1}$. Writing $\frac{1}{d_{o1}}+\frac{1}{d_{i1}}=\frac{1}{f_{1}}$ and $\frac{1}{-d_{i1}}+\frac{1}{d_{i2}}=\frac{1}{f_{2}}$, then adding side by side, the terms in $d_{i1}$ cancel: $\frac{1}{d_{o1}}+\frac{1}{d_{i2}}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$. The doublet therefore behaves like a single lens of power $P=P_{1}+P_{2}$. Here: $P=4.333+(-2.000)=2.333$ D, that is $f\approx 42.9$ cm. It is this simple fact that lets the optometrist assemble any correction by stacking trial lenses.

In both cases, the OBJECTIVE, facing the object, forms a first real image of it, called the intermediate image. This image then serves as the OBJECT for the EYEPIECE, which simply plays the role of a magnifying glass: positioned so that the intermediate image falls near its focal point, it produces a very enlarged virtual image, viewed by the eye. The total magnification is the product of the two effects. The difference between the two instruments lies in the object observed: close and tiny for the microscope, distant and large for the telescope.

The magnification of the objective is $m_{obj}=-\dfrac{d_{i}}{d_{o}}$. Since the object is placed just beyond the focal point, we use the usual approximation $d_{o}\approx f_{obj}$, hence $m_{obj}\approx-\dfrac{16.0}{0.800}=-20.0$. The eyepiece acts as a magnifying glass: $M_{eye}=\dfrac{25}{2.50}=10.0$. The total magnification is the product of the absolute values: $M=20.0\times 10.0=200$ times.

It is INVERTED. The objective gives an inverted image, since its magnification is negative, and the eyepiece, working as a magnifying glass, leaves it upright relative to itself: the final image is therefore inverted relative to the object. This causes no problem in microscopy: the observer simply moves the slide in the opposite direction to what is expected, which one gets used to within minutes. Adding an erecting system would cost light and sharpness for no benefit.

With the object at infinity, the objective forms the intermediate image exactly in its focal plane. For viewing with a relaxed eye, the eyepiece must have its own focal point at the same place: the two lenses are therefore separated by $f_{obj}+f_{eye}$. Comparing the angle at which the image is seen with the angle at which the object was seen gives $M=\dfrac{f_{obj}}{f_{eye}}$. Application: $M=\dfrac{120}{2.00}=60.0$ times.

Because magnification is easy to obtain but creates no information: changing the eyepiece is enough to double $M$, and all you then do is enlarge a blurry image. What really limits a telescope is the AMOUNT OF LIGHT collected, proportional to the area and so to the square of the diameter, and the RESOLVING POWER, the smallest angular detail that can be told apart, which is set by diffraction and also improves with diameter. An objective twice as wide collects four times as much light and resolves twice as finely: it shows objects that are genuinely fainter and finer. Magnification is useful only within the limit allowed by the diameter, beyond which we speak of empty magnification.

For a telescope, $M=\dfrac{f_{obj}}{f_{eye}}$: you need the LARGEST $f$ for the objective and the SMALLEST for the eyepiece. So take the 100 cm objective and the 2.00 cm eyepiece, giving $M=\dfrac{100}{2.00}=50.0$ times. The tube length is $f_{obj}+f_{eye}=100+2.00=102$ cm, an instrument a little over a metre long.

Swapping them gives $M=\dfrac{2.00}{100}=0.0200$, a magnification far below 1: the instrument REDUCES the image fifty times instead of enlarging it. You would have built a Galilean telescope backward, which is the principle of the wide-angle viewfinder. Swapping is therefore out of the question for observing the sky, and the rule is unambiguous: long focal length for the objective, short focal length for the eyepiece.

For a microscope, you need on the contrary an objective of VERY SHORT focal length, since the object is very close to it, and an eyepiece of moderate focal length. Take the 2.00 cm objective and the 5.00 cm eyepiece. The magnification of the objective is $\left|m_{obj}\right|\approx\dfrac{15.0}{2.00}=7.50$ and the magnification of the eyepiece is $\dfrac{25}{5.00}=5.00$. The total is $7.50\times 5.00=37.5$ times.

No, it cannot be used. As an objective, a focal length of 100 cm would force you to place the object more than a metre away, which is absurd for observing a microscope slide and would give no useful magnification. As an eyepiece, it would give a magnification of $\dfrac{25}{100}=0.25$, hence a REDUCTION. A long focal length converges weakly; yet the two stages of a microscope call for strong convergence, one to capture a tiny object placed very close, the other to act effectively as a magnifying glass.

For the TELESCOPE, you want an objective focal length as large as possible and an eyepiece focal length as small as possible, since $M$ is their ratio. For the MICROSCOPE, you want both focal lengths as small as possible, the objective to bring the object as close as possible and the eyepiece to magnify more, since $M$ is the product of two factors that are each inversely proportional to a focal length.